Element has Idempotent Power in Finite Semigroup

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Theorem

Let $\struct {S, \circ}$ be a finite semigroup.

For every element in $\struct {S, \circ}$, there is a power of that element which is idempotent.


That is:

$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$


Proof

From Finite Semigroup Equal Elements for Different Powers, we have:

$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$


Let $m > n$.

Let $n = k, m = k + l$.

Then:

$\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$


Now we show that:

$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$

That is, that $x^{k l}$ is idempotent.

First:

\(\displaystyle x^k\) \(=\) \(\displaystyle x^{k + l}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^k \circ x^l\) \(=\) \(\displaystyle x^{k + l} \circ x^l\) both sides ${} \circ x^l$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^{k + l}\) \(=\) \(\displaystyle x^{k + 2 l}\) Index Laws for Semigroup: Sum of Indices‎
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^k\) \(=\) \(\displaystyle x^{k + 2 l}\) as $x^k = x^{k + l}$


From here we can easily prove by induction that:

$\forall n \in \N: x^k = x^{k + n l}$


In particular:

$x^k = x^{k + k l} = x^{k \paren {l + 1} }$


There are two cases to consider:

$(1): \quad$ If $l = 1$, then $x^k = x^{k \paren {l + 1} } = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.
$(2): \quad$ If $l > 1$, then:
\(\displaystyle x^k\) \(=\) \(\displaystyle x^{k + k l}\)
\(\displaystyle \) \(=\) \(\displaystyle x^{k \paren {l + 1} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^k \circ x^{k \paren {l - 1} }\) \(=\) \(\displaystyle x^{k \paren {l + 1} } \circ x^{k \paren {l - 1} }\) both sides ${} \circ x^{k \paren {l - 1} }$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^{k + k l - k}\) \(=\) \(\displaystyle x^{kl + k + k l - k}\) Index Laws for Semigroup: Sum of Indices
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^{k l}\) \(=\) \(\displaystyle x^{k l + k l}\)
\(\displaystyle \) \(=\) \(\displaystyle x^{k l} \circ x^{k l}\)

and again, $x^{k l}$ is idempotent.

$\blacksquare$


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